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Cooling Tower

1. The exit temperature of the cooled water is 31.9°C. 2. To cool 1 kg of water from 52°C to 27°C requires 0.654 kg of air at a volume of 0.5709 m3. The maximum water flow that can be cooled by 142 m3/min of air is 248.7 kg/min. 3. The mass flow rate of air is 3.244 kg/s. The make-up water required is 0.1810 kg/s.

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Gianne Karl Almarines
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456 views6 pages

Cooling Tower

1. The exit temperature of the cooled water is 31.9°C. 2. To cool 1 kg of water from 52°C to 27°C requires 0.654 kg of air at a volume of 0.5709 m3. The maximum water flow that can be cooled by 142 m3/min of air is 248.7 kg/min. 3. The mass flow rate of air is 3.244 kg/s. The make-up water required is 0.1810 kg/s.

Uploaded by

Gianne Karl Almarines
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOC, PDF, TXT or read online on Scribd
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COOLING TOWERS

1. A mechanical-draft cooling tower receives 115 m3 per second of atmospheric air


at 103 kPa, 32 C dry bulb temperature, 55% RH and discharges the air saturated
at 36 C. If the tower receives 200 kg/s of water at 40 C, what will be the exit
temperature of the cooled water?

Solution:

m 3  200 kg s
t3  40 C

tdb2  36 C saturated

V1  115 m3 s
tdb1  32 C
1  55%

at 1, for tdb  32 C
1

pd  4.799 kPa
hg1  2559.9 kJ kg

ps  1 pd  0.55 4.799  2.639 kPa


0.622 ps 0.622 2.639
W1    0.0164 kg kg
pt  ps 103  2.639

h1  c ptdb1  W1hg  1.0062 32   0.0164 2559.9  74.2 kJ kg


COOLING TOWERS

v1 
RT

 0.287 32  273  0.8722 m3 kg
pt  ps 103  2.639
V 115
a  1 
m  131.85 kg s
v1 0.8722

at 2, tdb2  36 C , saturated
ps  pd  5.979 kPa
hg 2  2567.1 kJ kg

0.622 ps 0.622 5.979


W2    0.0383 kg kg
pt  ps 103  5.979

h2  c ptdb2  W2 hg  1.0062 36   0.0383 2567.1  134.5 kJ kg

At 3, t3  40 C
h3  h f at 40 C  167.57 kJ kg

To solve for m
4:
By Mass Balance:
3 m
m  a  W1m
a m
4 m
 a  W2 m
a
 3  W1  W2  m
4 m
m  a  200   0.0164  0.0383131.85  197.1 kg s

To solve for h4 :
 a h1  m
m  3h3  m
 4 h4  m
 a h2
131.85 74.2   200167.57   197.1 h4   131.85134.5
h4  129.70 kJ kg
 t4  31.9 C - exit water temperature.

2. In a cooling tower water enters at 52 C and leaves at 27 C. Air at 29 C and 47%


RH also enters the cooling tower and leaves at 46 C fully saturated with moisture.
It is desired to determine (a) the volume and mass of air necessary to cool 1 kg of
water, and (b) the quantity of water that can be cooled with 142 cu m per minute
of atmospheric air.

Solution:
COOLING TOWERS

t3  52 C
tdb2  46 C sat
tdb1  29 C
1  47%
t4  27 C

From psychrometric chart


At 1, tdb1  29 C , 1  47%
h1  59 kJ kg
W1  0.0116 kg kg
v1  0.873 m3 kg

at 2, tdb2  46 C , saturated
ps  pd  10.144 kPa
hg 2  2585.0 kJ kg
0.622 ps 0.62210.144
W2    0.0692 kg kg
pt  ps 101.325  10.144

h2  c ptdb2  W2 hg  1.0062 46   0.0692 2585  225.2 kJ kg

At 3, t3  52 C
h3  h f at 52 C  217.69 kJ kg

At 4, t4  27 C
h4  h f at 27 C  113 .25 kJ kg

(a) Volume of air necessary to cool 1 kg of water = V1  ma v1


COOLING TOWERS

To solve for ma when m3  1 kg

By energy balance:
Eq. (1) ma h1  m3h3  ma h2  m4 h4
By mass balance
m3  m4  ma W2  W1 
Eq. (2) m4  m3  ma W2  W1 
Substitute in (1)

Eq. (3) ma h1  m3h3  ma h2   m3  ma W2  W1   h4


 ma  59  1 217.69   ma  225.2  1   ma  0.0692  0.0116  113 .25
ma  0.654 kg

Volume of air = V1  ma v1   0.654  0.873  0.5709 m3

Mass of air required = ma  0.654 kg

(b) Let m
 3 = quantity of water.
V1 142
a 
m   162.66 kg min
v1 0.873

Use Eq. (3), change ma  m a , m3  m 3


m a h1  m 3h3  m a h2   m 3  m a W2  W1   h4
162.66 59   m 3  217.69  162.66 225.2   m 3  162.66 0.0692  0.0116  113 .25
9597  217.69m  3  36,631  113.25m
 3  1061.1
 3  248.7 kg min
m

 Quantity of water =  3  248.7 kg min


m

3. A cooling tower receives 6 kg/s of water of 60 C. Air enters the tower at 32 C dry
bulb and 27 C wet bulb temperatures and leaves at 50 C and 90% relative
humidity. The cooling efficiency is 60.6%. Determine (a) the mass flow rate of air
entering, and (b) the quantity of make-up water required.

Solution:
COOLING TOWERS

 3  6 kg s , t3  60 C
m
t db2  50 C
2  90%
tdb1  32 C
t wb1  27 C

Cooling tower efficiency = 60.6%

To solve for t 4 :
t3  t 4
Efficiency 
t3  t wb1
60  t4
0.606 
60  27
t3  40 C

at 1, tdb  32 C , t wb  27 C
1 1

h1  85 kJ kg
W1  0.0208 kg kg

at 2, t db2  50 C , 2  90%
pd  12.349 kPa
ps  2 pd   0.9012.349  11 .114 kPa
hg 2  2592.1 kJ kg
0.622 ps 0.62211 .114 
W2    0.0766 kg kg
pt  ps 101.325  11.114

h2  c ptdb2  W2 hg  1.0062 50   0.0766 2592.1  248.9 kJ kg


COOLING TOWERS

At 3, t3  60 C
h3  h f at 60 C  251.13 kJ kg

At 4, t4  40 C
h4  h f at 40 C  167.57 kJ kg
By mass balance
m3  m4  ma W2  W1 
m4  m3  ma W2  W1 

By energy valance
 a h1  m
m  3h3  m
 a h2  m
 4 h4
m a h1  m 3h3  m a h2   m 3  m a W2  W1   h4
 m a  85   6 251.13   m a  248.9   6   m a  0.0766  0.0208 167.57
85m  a  1506.8  248.9m
 a  1005.4  9.35m
a
 a  3.244 kg s
m

(a) Mass flow rate of air = m a  3.244 kg s


(b) Make Up Water = ma W2  W1    3.244 0.0766  0.0208  0.1810 kg s

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